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Chemistry Question on Order of Reaction

The rate of the reaction A→A\xrightarrow{{}} products, at the initial concentration of 3.24×10−2M3.24\times {{10}^{-2}}M is nine times its rate at another initial concentration of 1.2×10−3M1.2\times {{10}^{-3}}M . The order of the reaction is

A

12\frac{1}{2}

B

34\frac{3}{4}

C

32\frac{3}{2}

D

23\frac{2}{3}

Answer

23\frac{2}{3}

Explanation

Solution

For the reaction A→A\xrightarrow[{}]{{}} products; let the order of reaction is n. Then r=k[A]nr=k{{[A]}^{n}} =k×(1.2×10−3)n=k\times {{(1.2\times {{10}^{-3}})}^{n}} ...(i) At the initial concentration of 3.24×10−2M,3.24\times {{10}^{-2}}M, the rate is θ\theta times. Hence, 9r=k×(3.24×10−2)n9r=k\times {{(3.24\times {{10}^{-2}})}^{n}} ...(ii) On dividing, 9rr=k×(3.24×10−2)nk×(1.2×10−3)n\frac{9r}{r}=\frac{k\times {{(3.24\times {{10}^{-2}})}^{n}}}{k\times {{(1.2\times {{10}^{-3}})}^{n}}} 9=(27)n9={{(27)}^{n}} 9=(33)2/39={{({{3}^{3}})}^{2/3}} ∴\therefore n=23n=\frac{2}{3}